A 0.50 mol sample of an ideal gas at 1.00 atm and 25°C occupies what volume? Use R = 0.0821 L·atm/mol·K.

A 0.50 mol sample of an ideal gas at 1.00 atm and 25°C occupies what volume? Use R = 0.0821 L·atm/mol·K.

• A. 12.23 L
• B. 24.47 L
• C. 6.02 L
• D. 48.0 L

Answer: (A)

The main idea is using the ideal gas law, PV = nRT, to relate how much gas, its temperature, and the pressure determine the volume. Solve for V: V = nRT / P. Convert 25°C to kelvin: 25 + 273.15 = 298 K. Then V = (0.50 mol)(0.0821 L·atm/mol·K)(298 K) / (1.00 atm) ≈ 12.23 L. So, at 25°C and 1 atm, half a mole of an ideal gas occupies about 12.23 liters. For context, 1.00 mole would be about 24.47 L, illustrating how volume scales with the amount of gas. The other numbers would require different amounts of gas or different conditions to match those volumes.

The main idea is using the ideal gas law, PV = nRT, to relate how much gas, its temperature, and the pressure determine the volume. Solve for V: V = nRT / P. Convert 25°C to kelvin: 25 + 273.15 = 298 K. Then V = (0.50 mol)(0.0821 L·atm/mol·K)(298 K) / (1.00 atm) ≈ 12.23 L. So, at 25°C and 1 atm, half a mole of an ideal gas occupies about 12.23 liters. For context, 1.00 mole would be about 24.47 L, illustrating how volume scales with the amount of gas. The other numbers would require different amounts of gas or different conditions to match those volumes.